PAPER · REASONING · PRACTICE

Angle bisectors: the converse and applications — Teacher reference

Separate the two directions, construct an HL proof, and use equal distances in angle and perimeter problems.

40 minutes · Year 8, upper volume · text pp. 27–28 · foundation lesson 14

Textbook reference: Year 8, upper volume · congruent triangles · angle bisectors

One worked example, one guided attempt, eight core questions, four follow-ups and two extensions. Write on paper; the teacher controls the reveal.

Student classroom page Arrange in My Classroom

Lesson routePhase

4′Check prerequisites
5′Observe and predict
8′Worked example and guided attempt
15′Independent practice
5′Discuss and correct
3′Exit questions

Observe · explain

A question before practising

Hide the angle-equality marks. If the perpendicular lengths match, which triangles and which given sides can locate the bisector?

Organise the key idea

An interior point with equal perpendicular distances to the sides lies on the internal bisector. The equal distances are inputs here, whereas they were the goal of the forward proof.

Open the original chapter Open the interactive inquiry

Use the lesson question here for paper discussion; the original activity opens separately.

Worked example1

P is inside ∠AOB. PM⊥OA, PN⊥OB and PM=PN. Prove that OP bisects ∠AOB.

Geometry diagram; not drawn to scale except for construction tasks.

Identify the right angle in each triangle and the opposite side OP.
For HL, use common hypotenuse OP and the given equal legs PM, PN.
Congruence gives equal angles at O; retain the condition that P is inside the angle.

The perpendicular conditions make △OPM and △OPN right triangles.
OP=OP is the common hypotenuse, and PM=PN is given.
Thus △OPM≅△OPN by HL.
Corresponding angles give ∠MOP=∠PON.
Since P is inside the angle, OP is its internal bisector.

Check your work

State both right triangles and identify the common hypotenuse.
Use the given equal legs and HL.
Deduce equal vertex angles and retain the interior condition.
Accept another complete proof if every condition and deduction is justified.

Guided attempt2

P is inside ∠AOB, PM⊥OA, PN⊥OB and PM=PN. Complete the HL proof without assuming any equal angles at O.

The two right angles are at ___ and ___.
Hypotenuse ___=___; leg ___=___ is given.
△OPM≅△OPN by ___, so ∠___=∠___.
Because P is ___, OP is the ___ bisector.

Geometry diagram; not drawn to scale except for construction tasks.

Identify the right angle in each triangle and the opposite side OP.
For HL, use common hypotenuse OP and the given equal legs PM, PN.
Congruence gives equal angles at O; retain the condition that P is inside the angle.

The perpendicular conditions make △OPM and △OPN right triangles.
OP=OP is the common hypotenuse, and PM=PN is given.
Thus △OPM≅△OPN by HL.
Corresponding angles give ∠MOP=∠PON.
Since P is inside the angle, OP is its internal bisector.

Check your work

State both right triangles and identify the common hypotenuse.
Use the given equal legs and HL.
Deduce equal vertex angles and retain the interior condition.
Accept another complete proof if every condition and deduction is justified.

Prerequisite3

PM⊥OA and M lies on OA. Which side is the hypotenuse of △OPM?

OM
Any side can be called the hypotenuse.
OP
PM

Geometry diagram; not drawn to scale except for construction tasks.

Locate the right angle first.
The hypotenuse does not touch the right-angle vertex.
The right angle is at M, so its opposite side is OP.

PM⊥OM gives a right angle at M.
The side opposite M is OP.

Check your work

Identify OP by its position opposite the right angle.

Prerequisite4

Which statement is the converse of “a point on the internal bisector has equal perpendicular distances to the sides”?

Every interior point has equal distances to the sides.
A point on a side is the angle vertex.
Two equal angles always have equal side lengths.
An interior point with equal perpendicular distances lies on the internal bisector.

Identify what was given in the forward statement.
Equal distances become the given facts in the converse.
The converse concludes that the interior point lies on the bisector.

Exchange the original premise and conclusion.
Keep the domain restriction: P must be inside the angle.

Check your work

Reverse the logical direction without deleting necessary conditions.

Independent practice5

P is inside ∠AOB=86°. PM⊥OA, PN⊥OB and PM=PN=4 cm. Find ∠AOP and give the reason for halving.

Geometry diagram; not drawn to scale except for construction tasks.

You have not been given that OP is a bisector.
Use equal perpendicular distances to establish the bisector first.
After that, halve 86°.

The equal perpendicular distances and interior condition give the internal bisector.
Thus ∠AOP=86°÷2=43°.

Check your work

State the converse before the numerical halving.

Independent practice6

P is inside ∠AOB. PM⊥OA, PN⊥OB and PM=PN. If ∠MOP=28°, find ∠AOB.

Geometry diagram; not drawn to scale except for construction tasks.

Use the converse to find the other angle at O.
The two parts are equal.
Add two copies of 28°.

The converse gives ∠MOP=∠PON.
The whole angle is 28°+28°=56°.

Check your work

Justify equal angles and sum them correctly.

Independent practice7

Distances to the two intersecting lines containing OA and OB are equal, but P is not stated to be inside ∠AOB. Can we conclude that P lies on this angle’s internal bisector?

Yes, because all intersecting lines are perpendicular.
No point can have two equal distances.
Not necessarily; the interior condition is needed for that conclusion.
Yes, with no position condition.

Check the domain written in the theorem.
Two intersecting full lines form several angles.
Retain “P is inside ∠AOB” when using the stated converse.

The proposed conclusion concerns a particular internal angle.
An equidistant point outside that angle need not lie on its internal bisector.

Check your work

Explain the role of the interior restriction.

Independent practice8

X is inside ∠AVB. XH⊥VA, XK⊥VB and XH=XK. Prove that VX is the internal angle bisector.

Geometry diagram; not drawn to scale except for construction tasks.

Identify the right angle in each triangle and the opposite side VX.
For HL, use common hypotenuse VX and the given equal legs XH, XK.
Congruence gives equal angles at V; retain the condition that X is inside the angle.

The perpendicular conditions make △VXH and △VXK right triangles.
VX=VX is the common hypotenuse, and XH=XK is given.
Thus △VXH≅△VXK by HL.
Corresponding angles give ∠HVX=∠XVK.
Since X is inside the angle, VX is its internal bisector.

Check your work

State both right triangles and identify the common hypotenuse.
Use the given equal legs and HL.
Deduce equal vertex angles and retain the interior condition.
Accept another complete proof if every condition and deduction is justified.

Exit question9

P is inside ∠AOB. M lies on OA, N lies on OB and PM=PN. No perpendicular conditions are given. Is this enough to prove that OP is the bisector?

Check the definition of distance.
M and N being on the sides does not make the segments perpendicular.
The missing facts are PM⊥OA and PN⊥OB.

The converse compares distances, meaning perpendicular lengths.
Equal arbitrary segments ending on the sides do not supply that premise.

Check your work

Reject the inference and name the missing perpendicular facts.

Exit question10

P is inside the angle and its two perpendicular distances are equal. Which result directly tells us where P lies?

Equality of all three angles, without side data.
The converse of the angle-bisector property.
The forward property, starting by assuming OP bisects the angle.
The triangle perimeter formula alone.

Separate given facts from the target.
The target is not a length here.
Equal distances imply the interior bisector by the converse.

Equal perpendicular distances are the given facts.
The desired conclusion is the bisector, so use the converse.

Check your work

Choose the direction that matches the given facts.

Further practice11

P lies on the internal bisector, PM⊥OA and PN⊥OB. PM=3 cm and △OPM has perimeter 12 cm. Find OP+ON.

Geometry diagram; not drawn to scale except for construction tasks.

Find the corresponding pair involving ON.
OM=ON, so subtract PM from the perimeter.
Compute 12−3.

The right triangles are congruent, giving OM=ON.
OP+ON=OP+OM=12−3=9 cm.

Check your work

Use OM=ON and subtract the given leg.

Further practice12

P lies on the internal bisector, PM⊥OA and PN⊥OB. PM=9 cm and △OPM has perimeter 36 cm. Find OP+ON.

Geometry diagram; not drawn to scale except for construction tasks.

The perimeter contains OP, OM and PM.
Replace ON by OM.
Subtract PM=9 from 36.

Congruence gives OM=ON.
The two requested sides total 36−9=27 cm.

Check your work

Identify the correct corresponding side and total.

Further practice13

P is inside the angle, PM⊥OA, PN⊥OB and PM=PN=5 cm. Which conclusion follows directly?

OP bisects the angle; its length is not determined by 5 cm alone.
OP=5 cm.
The whole angle must be 90°.
M=N.

Geometry diagram; not drawn to scale except for construction tasks.

Distinguish the hypotenuse from a perpendicular leg.
The theorem concludes an angle bisector.
No angle size or OP length has been supplied.

Equal perpendicular distances satisfy the converse.
Different angle openings can have that same perpendicular distance but different OP lengths.

Check your work

State the bisector conclusion without inventing a length or angle.

Further practice14

Complete two sentences: forward direction, given ___, prove equal perpendicular distances; converse, given ___, prove the internal bisector. Include all position and perpendicular conditions.

Ask what is known before any proof begins.
The roles of bisector and equal distances exchange.
Write the necessary domain and perpendicular facts explicitly.

In the forward proof, the bisector gives equal vertex angles.
In the converse, equal legs and the common hypotenuse give HL.
Keep “inside” and “perpendicular distance” in the converse.

Check your work

Give both directions without circular assumptions.

Extension15

I is inside △ABC and lies on the internal bisectors at A and B. Let D,E,F be its perpendicular feet on AB,AC,BC respectively. Prove that I also lies on the internal bisector at C.

Write two equalities involving the distance to AB.
ID links IE and IF.
Use IE=IF with the interior condition at C.

The bisector at A gives ID=IE.
The bisector at B gives ID=IF.
Therefore IE=IF. These are perpendicular distances to CA and CB.
I is inside ∠ACB, so its converse gives the third bisector.

Check your work

Use both forward properties and the final converse with the interior condition.
Accept another complete proof if every condition and deduction is justified.

Extension16

Someone puts P=O, obtains two zero distances, and tries to use HL to define the ray OP as an angle bisector. Explain why this proof setup fails.

Can two identical points determine a ray?
Check whether the two triangles have positive area.
The vertex is not an interior point of the open angle.

A ray OP requires P≠O.
With P=O there are no nondegenerate right triangles for HL.
Use the interior-point hypothesis as stated.

Check your work

Identify both the undefined ray and degenerate-triangle problems.

Preparation and assessment notes

Prerequisites

The forward property, right-triangle hypotenuse and leg, HL, and the meaning of a point inside an angle.

Look for in students’ work

Model HL with OP as the hypotenuse, then ask pupils to label given and required facts separately. Make the interior and perpendicular conditions visible in each proof.

Collect the paper exit questions and note completion and common errors. This page does not grade or store paper answers.

All reference answers

Worked example1

P is inside ∠AOB. PM⊥OA, PN⊥OB and PM=PN. Prove that OP bisects ∠AOB.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

The right triangles are congruent by HL, so the vertex angles match and OP is the internal bisector.

The perpendicular conditions make △OPM and △OPN right triangles.
OP=OP is the common hypotenuse, and PM=PN is given.
Thus △OPM≅△OPN by HL.
Corresponding angles give ∠MOP=∠PON.
Since P is inside the angle, OP is its internal bisector.

State both right triangles and identify the common hypotenuse.
Use the given equal legs and HL.
Deduce equal vertex angles and retain the interior condition.
Accept another complete proof if every condition and deduction is justified.

Guided attempt2

P is inside ∠AOB, PM⊥OA, PN⊥OB and PM=PN. Complete the HL proof without assuming any equal angles at O.

The two right angles are at ___ and ___.
Hypotenuse ___=___; leg ___=___ is given.
△OPM≅△OPN by ___, so ∠___=∠___.
Because P is ___, OP is the ___ bisector.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

OP is the common hypotenuse; PM=PN are equal legs; HL gives ∠MOP=∠PON, hence the internal bisector.

The perpendicular conditions make △OPM and △OPN right triangles.
OP=OP is the common hypotenuse, and PM=PN is given.
Thus △OPM≅△OPN by HL.
Corresponding angles give ∠MOP=∠PON.
Since P is inside the angle, OP is its internal bisector.

State both right triangles and identify the common hypotenuse.
Use the given equal legs and HL.
Deduce equal vertex angles and retain the interior condition.
Accept another complete proof if every condition and deduction is justified.

Prerequisite3

PM⊥OA and M lies on OA. Which side is the hypotenuse of △OPM?

OM
Any side can be called the hypotenuse.
OP
PM

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

OP, because it is opposite the right angle at M.

PM⊥OM gives a right angle at M.
The side opposite M is OP.

Identify OP by its position opposite the right angle.

Prerequisite4

Which statement is the converse of “a point on the internal bisector has equal perpendicular distances to the sides”?

Every interior point has equal distances to the sides.
A point on a side is the angle vertex.
Two equal angles always have equal side lengths.
An interior point with equal perpendicular distances lies on the internal bisector.

Reference answer

The first statement, retaining the interior and perpendicular-distance conditions.

Exchange the original premise and conclusion.
Keep the domain restriction: P must be inside the angle.

Reverse the logical direction without deleting necessary conditions.

Independent practice5

P is inside ∠AOB=86°. PM⊥OA, PN⊥OB and PM=PN=4 cm. Find ∠AOP and give the reason for halving.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

43°: the converse first proves that OP bisects the angle.

The equal perpendicular distances and interior condition give the internal bisector.
Thus ∠AOP=86°÷2=43°.

State the converse before the numerical halving.

Independent practice6

P is inside ∠AOB. PM⊥OA, PN⊥OB and PM=PN. If ∠MOP=28°, find ∠AOB.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

∠AOB=56°.

The converse gives ∠MOP=∠PON.
The whole angle is 28°+28°=56°.

Justify equal angles and sum them correctly.

Independent practice7

Distances to the two intersecting lines containing OA and OB are equal, but P is not stated to be inside ∠AOB. Can we conclude that P lies on this angle’s internal bisector?

Yes, because all intersecting lines are perpendicular.
No point can have two equal distances.
Not necessarily; the interior condition is needed for that conclusion.
Yes, with no position condition.

Reference answer

Not necessarily. Equal distances to two full lines also occur on the other bisector.

The proposed conclusion concerns a particular internal angle.
An equidistant point outside that angle need not lie on its internal bisector.

Explain the role of the interior restriction.

Independent practice8

X is inside ∠AVB. XH⊥VA, XK⊥VB and XH=XK. Prove that VX is the internal angle bisector.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

△VXH≅△VXK by HL; their equal angles at V make VX the internal bisector.

The perpendicular conditions make △VXH and △VXK right triangles.
VX=VX is the common hypotenuse, and XH=XK is given.
Thus △VXH≅△VXK by HL.
Corresponding angles give ∠HVX=∠XVK.
Since X is inside the angle, VX is its internal bisector.

State both right triangles and identify the common hypotenuse.
Use the given equal legs and HL.
Deduce equal vertex angles and retain the interior condition.
Accept another complete proof if every condition and deduction is justified.

Exit question9

P is inside ∠AOB. M lies on OA, N lies on OB and PM=PN. No perpendicular conditions are given. Is this enough to prove that OP is the bisector?

Reference answer

No. PM and PN need not be the perpendicular distances to the sides.

The converse compares distances, meaning perpendicular lengths.
Equal arbitrary segments ending on the sides do not supply that premise.

Reject the inference and name the missing perpendicular facts.

Exit question10

P is inside the angle and its two perpendicular distances are equal. Which result directly tells us where P lies?

Equality of all three angles, without side data.
The converse of the angle-bisector property.
The forward property, starting by assuming OP bisects the angle.
The triangle perimeter formula alone.

Reference answer

The converse places P on the internal angle bisector.

Equal perpendicular distances are the given facts.
The desired conclusion is the bisector, so use the converse.

Choose the direction that matches the given facts.

Further practice11

P lies on the internal bisector, PM⊥OA and PN⊥OB. PM=3 cm and △OPM has perimeter 12 cm. Find OP+ON.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

OP+ON=9 cm.

The right triangles are congruent, giving OM=ON.
OP+ON=OP+OM=12−3=9 cm.

Use OM=ON and subtract the given leg.

Further practice12

P lies on the internal bisector, PM⊥OA and PN⊥OB. PM=9 cm and △OPM has perimeter 36 cm. Find OP+ON.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

OP+ON=27 cm.

Congruence gives OM=ON.
The two requested sides total 36−9=27 cm.

Identify the correct corresponding side and total.

Further practice13

P is inside the angle, PM⊥OA, PN⊥OB and PM=PN=5 cm. Which conclusion follows directly?

OP bisects the angle; its length is not determined by 5 cm alone.
OP=5 cm.
The whole angle must be 90°.
M=N.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

The first conclusion: the converse determines the direction, not OP’s length from one distance alone.

Equal perpendicular distances satisfy the converse.
Different angle openings can have that same perpendicular distance but different OP lengths.

State the bisector conclusion without inventing a length or angle.

Further practice14

Complete two sentences: forward direction, given ___, prove equal perpendicular distances; converse, given ___, prove the internal bisector. Include all position and perpendicular conditions.

Reference answer

Forward: an interior point on the bisector, with perpendicular feet. Converse: an interior point with equal perpendicular distances to the sides.

In the forward proof, the bisector gives equal vertex angles.
In the converse, equal legs and the common hypotenuse give HL.
Keep “inside” and “perpendicular distance” in the converse.

Give both directions without circular assumptions.

Extension15

I is inside △ABC and lies on the internal bisectors at A and B. Let D,E,F be its perpendicular feet on AB,AC,BC respectively. Prove that I also lies on the internal bisector at C.

Reference answer

ID=IE and ID=IF, hence IE=IF. The interior converse at C places I on C’s bisector.

The bisector at A gives ID=IE.
The bisector at B gives ID=IF.
Therefore IE=IF. These are perpendicular distances to CA and CB.
I is inside ∠ACB, so its converse gives the third bisector.

Use both forward properties and the final converse with the interior condition.
Accept another complete proof if every condition and deduction is justified.

Extension16

Someone puts P=O, obtains two zero distances, and tries to use HL to define the ray OP as an angle bisector. Explain why this proof setup fails.

Reference answer

OP is not a ray determined by two distinct points, and both supposed triangles are degenerate. The theorem’s interior-point setup excludes the vertex.

A ray OP requires P≠O.
With P=O there are no nondegenerate right triangles for HL.
Use the interior-point hypothesis as stated.

Identify both the undefined ray and degenerate-triangle problems.

Paper for the class

Download all worksheets (ZIP · 318 PDFs · 47.2 MB)

All 53 Year 7–8 upper- and lower-volume packs in Traditional Chinese and English. Includes teacher answers; student core, further practice and teacher papers are in separate folders.

Core student worksheet: two A4 pages. Further practice is separate. Teacher copies include reference reasoning.

Print edition

Textbook reference and authorship

Original practice aligned to the referenced skills. Use your own textbook for the cited pages.

大湾区初中数学三维课堂（八年级上册）正文.pdf

Year 8, upper volume · text pp. 27–28 · foundation lesson 14

Original questions and explanations; the source textbook and answer PDFs are not reproduced here.

Angle bisectors: the converse and applications

Year 8, upper volume · text pp. 27–28 · foundation lesson 14 · 1/2

Name: __________ Class: ______ Date: __________

Worked example1

P is inside ∠AOB. PM⊥OA, PN⊥OB and PM=PN. Prove that OP bisects ∠AOB.

Geometry diagram; not drawn to scale except for construction tasks.

Guided attempt2

P is inside ∠AOB, PM⊥OA, PN⊥OB and PM=PN. Complete the HL proof without assuming any equal angles at O.

The two right angles are at ___ and ___.
Hypotenuse ___=___; leg ___=___ is given.
△OPM≅△OPN by ___, so ∠___=∠___.
Because P is ___, OP is the ___ bisector.

Geometry diagram; not drawn to scale except for construction tasks.

Prerequisite3

PM⊥OA and M lies on OA. Which side is the hypotenuse of △OPM?

OM
Any side can be called the hypotenuse.
OP
PM

Geometry diagram; not drawn to scale except for construction tasks.

Prerequisite4

Which statement is the converse of “a point on the internal bisector has equal perpendicular distances to the sides”?

Every interior point has equal distances to the sides.
A point on a side is the angle vertex.
Two equal angles always have equal side lengths.
An interior point with equal perpendicular distances lies on the internal bisector.

Angle bisectors: the converse and applications

Year 8, upper volume · text pp. 27–28 · foundation lesson 14 · 2/2

Name: __________ Class: ______ Date: __________

Independent practice5

P is inside ∠AOB=86°. PM⊥OA, PN⊥OB and PM=PN=4 cm. Find ∠AOP and give the reason for halving.

Geometry diagram; not drawn to scale except for construction tasks.

Independent practice6

P is inside ∠AOB. PM⊥OA, PN⊥OB and PM=PN. If ∠MOP=28°, find ∠AOB.

Geometry diagram; not drawn to scale except for construction tasks.

Independent practice7

Distances to the two intersecting lines containing OA and OB are equal, but P is not stated to be inside ∠AOB. Can we conclude that P lies on this angle’s internal bisector?

Yes, because all intersecting lines are perpendicular.
No point can have two equal distances.
Not necessarily; the interior condition is needed for that conclusion.
Yes, with no position condition.

Independent practice8

X is inside ∠AVB. XH⊥VA, XK⊥VB and XH=XK. Prove that VX is the internal angle bisector.

Geometry diagram; not drawn to scale except for construction tasks.

Exit question9

P is inside ∠AOB. M lies on OA, N lies on OB and PM=PN. No perpendicular conditions are given. Is this enough to prove that OP is the bisector?

Exit question10

P is inside the angle and its two perpendicular distances are equal. Which result directly tells us where P lies?

Equality of all three angles, without side data.
The converse of the angle-bisector property.
The forward property, starting by assuming OP bisects the angle.
The triangle perimeter formula alone.

Angle bisectors: the converse and applications — Further practice

Year 8, upper volume · text pp. 27–28 · foundation lesson 14

Further practice11

P lies on the internal bisector, PM⊥OA and PN⊥OB. PM=3 cm and △OPM has perimeter 12 cm. Find OP+ON.

Geometry diagram; not drawn to scale except for construction tasks.

Further practice12

P lies on the internal bisector, PM⊥OA and PN⊥OB. PM=9 cm and △OPM has perimeter 36 cm. Find OP+ON.

Geometry diagram; not drawn to scale except for construction tasks.

Further practice13

P is inside the angle, PM⊥OA, PN⊥OB and PM=PN=5 cm. Which conclusion follows directly?

OP bisects the angle; its length is not determined by 5 cm alone.
OP=5 cm.
The whole angle must be 90°.
M=N.

Geometry diagram; not drawn to scale except for construction tasks.

Further practice14

Complete two sentences: forward direction, given ___, prove equal perpendicular distances; converse, given ___, prove the internal bisector. Include all position and perpendicular conditions.

Extension15

I is inside △ABC and lies on the internal bisectors at A and B. Let D,E,F be its perpendicular feet on AB,AC,BC respectively. Prove that I also lies on the internal bisector at C.

Extension16

Someone puts P=O, obtains two zero distances, and tries to use HL to define the ray OP as an angle bisector. Explain why this proof setup fails.

Angle bisectors: the converse and applications — Teacher answers

Year 8, upper volume · text pp. 27–28 · foundation lesson 14

Model HL with OP as the hypotenuse, then ask pupils to label given and required facts separately. Make the interior and perpendicular conditions visible in each proof.

Worked example1

P is inside ∠AOB. PM⊥OA, PN⊥OB and PM=PN. Prove that OP bisects ∠AOB.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

The right triangles are congruent by HL, so the vertex angles match and OP is the internal bisector.

The perpendicular conditions make △OPM and △OPN right triangles.
OP=OP is the common hypotenuse, and PM=PN is given.
Thus △OPM≅△OPN by HL.
Corresponding angles give ∠MOP=∠PON.
Since P is inside the angle, OP is its internal bisector.

State both right triangles and identify the common hypotenuse.
Use the given equal legs and HL.
Deduce equal vertex angles and retain the interior condition.
Accept another complete proof if every condition and deduction is justified.

Guided attempt2

P is inside ∠AOB, PM⊥OA, PN⊥OB and PM=PN. Complete the HL proof without assuming any equal angles at O.

The two right angles are at ___ and ___.
Hypotenuse ___=___; leg ___=___ is given.
△OPM≅△OPN by ___, so ∠___=∠___.
Because P is ___, OP is the ___ bisector.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

OP is the common hypotenuse; PM=PN are equal legs; HL gives ∠MOP=∠PON, hence the internal bisector.

The perpendicular conditions make △OPM and △OPN right triangles.
OP=OP is the common hypotenuse, and PM=PN is given.
Thus △OPM≅△OPN by HL.
Corresponding angles give ∠MOP=∠PON.
Since P is inside the angle, OP is its internal bisector.

State both right triangles and identify the common hypotenuse.
Use the given equal legs and HL.
Deduce equal vertex angles and retain the interior condition.
Accept another complete proof if every condition and deduction is justified.

Prerequisite3

PM⊥OA and M lies on OA. Which side is the hypotenuse of △OPM?

OM
Any side can be called the hypotenuse.
OP
PM

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

OP, because it is opposite the right angle at M.

PM⊥OM gives a right angle at M.
The side opposite M is OP.

Identify OP by its position opposite the right angle.

Prerequisite4

Which statement is the converse of “a point on the internal bisector has equal perpendicular distances to the sides”?

Every interior point has equal distances to the sides.
A point on a side is the angle vertex.
Two equal angles always have equal side lengths.
An interior point with equal perpendicular distances lies on the internal bisector.

Reference answer

The first statement, retaining the interior and perpendicular-distance conditions.

Exchange the original premise and conclusion.
Keep the domain restriction: P must be inside the angle.

Reverse the logical direction without deleting necessary conditions.

Independent practice5

P is inside ∠AOB=86°. PM⊥OA, PN⊥OB and PM=PN=4 cm. Find ∠AOP and give the reason for halving.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

43°: the converse first proves that OP bisects the angle.

The equal perpendicular distances and interior condition give the internal bisector.
Thus ∠AOP=86°÷2=43°.

State the converse before the numerical halving.

Independent practice6

P is inside ∠AOB. PM⊥OA, PN⊥OB and PM=PN. If ∠MOP=28°, find ∠AOB.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

∠AOB=56°.

The converse gives ∠MOP=∠PON.
The whole angle is 28°+28°=56°.

Justify equal angles and sum them correctly.

Independent practice7

Distances to the two intersecting lines containing OA and OB are equal, but P is not stated to be inside ∠AOB. Can we conclude that P lies on this angle’s internal bisector?

Yes, because all intersecting lines are perpendicular.
No point can have two equal distances.
Not necessarily; the interior condition is needed for that conclusion.
Yes, with no position condition.

Reference answer

Not necessarily. Equal distances to two full lines also occur on the other bisector.

The proposed conclusion concerns a particular internal angle.
An equidistant point outside that angle need not lie on its internal bisector.

Explain the role of the interior restriction.

Independent practice8

X is inside ∠AVB. XH⊥VA, XK⊥VB and XH=XK. Prove that VX is the internal angle bisector.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

△VXH≅△VXK by HL; their equal angles at V make VX the internal bisector.

The perpendicular conditions make △VXH and △VXK right triangles.
VX=VX is the common hypotenuse, and XH=XK is given.
Thus △VXH≅△VXK by HL.
Corresponding angles give ∠HVX=∠XVK.
Since X is inside the angle, VX is its internal bisector.

State both right triangles and identify the common hypotenuse.
Use the given equal legs and HL.
Deduce equal vertex angles and retain the interior condition.
Accept another complete proof if every condition and deduction is justified.

Exit question9

P is inside ∠AOB. M lies on OA, N lies on OB and PM=PN. No perpendicular conditions are given. Is this enough to prove that OP is the bisector?

Reference answer

No. PM and PN need not be the perpendicular distances to the sides.

The converse compares distances, meaning perpendicular lengths.
Equal arbitrary segments ending on the sides do not supply that premise.

Reject the inference and name the missing perpendicular facts.

Exit question10

P is inside the angle and its two perpendicular distances are equal. Which result directly tells us where P lies?

Equality of all three angles, without side data.
The converse of the angle-bisector property.
The forward property, starting by assuming OP bisects the angle.
The triangle perimeter formula alone.

Reference answer

The converse places P on the internal angle bisector.

Equal perpendicular distances are the given facts.
The desired conclusion is the bisector, so use the converse.

Choose the direction that matches the given facts.

Further practice11

P lies on the internal bisector, PM⊥OA and PN⊥OB. PM=3 cm and △OPM has perimeter 12 cm. Find OP+ON.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

OP+ON=9 cm.

The right triangles are congruent, giving OM=ON.
OP+ON=OP+OM=12−3=9 cm.

Use OM=ON and subtract the given leg.

Further practice12

P lies on the internal bisector, PM⊥OA and PN⊥OB. PM=9 cm and △OPM has perimeter 36 cm. Find OP+ON.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

OP+ON=27 cm.

Congruence gives OM=ON.
The two requested sides total 36−9=27 cm.

Identify the correct corresponding side and total.

Further practice13

P is inside the angle, PM⊥OA, PN⊥OB and PM=PN=5 cm. Which conclusion follows directly?

OP bisects the angle; its length is not determined by 5 cm alone.
OP=5 cm.
The whole angle must be 90°.
M=N.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

The first conclusion: the converse determines the direction, not OP’s length from one distance alone.

Equal perpendicular distances satisfy the converse.
Different angle openings can have that same perpendicular distance but different OP lengths.

State the bisector conclusion without inventing a length or angle.

Further practice14

Complete two sentences: forward direction, given ___, prove equal perpendicular distances; converse, given ___, prove the internal bisector. Include all position and perpendicular conditions.

Reference answer

Forward: an interior point on the bisector, with perpendicular feet. Converse: an interior point with equal perpendicular distances to the sides.

In the forward proof, the bisector gives equal vertex angles.
In the converse, equal legs and the common hypotenuse give HL.
Keep “inside” and “perpendicular distance” in the converse.

Give both directions without circular assumptions.

Extension15

I is inside △ABC and lies on the internal bisectors at A and B. Let D,E,F be its perpendicular feet on AB,AC,BC respectively. Prove that I also lies on the internal bisector at C.

Reference answer

ID=IE and ID=IF, hence IE=IF. The interior converse at C places I on C’s bisector.

The bisector at A gives ID=IE.
The bisector at B gives ID=IF.
Therefore IE=IF. These are perpendicular distances to CA and CB.
I is inside ∠ACB, so its converse gives the third bisector.

Use both forward properties and the final converse with the interior condition.
Accept another complete proof if every condition and deduction is justified.

Extension16

Someone puts P=O, obtains two zero distances, and tries to use HL to define the ray OP as an angle bisector. Explain why this proof setup fails.

Reference answer

OP is not a ray determined by two distinct points, and both supposed triangles are degenerate. The theorem’s interior-point setup excludes the vertex.

A ray OP requires P≠O.
With P=O there are no nondegenerate right triangles for HL.
Use the interior-point hypothesis as stated.

Identify both the undefined ray and degenerate-triangle problems.